Final Answer:
The correct solution to the equation
.
thus the correct option is (a).
Step-by-step explanation:
To find the solution, we'll first rearrange the equation to set it equal to zero:
![\[3v^2 + 17v + 24 - 5 = 0.\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/hc0umuzb384qm6h47137cp2n2k5xjhdfu6.png)
Combine like terms:
![\[3v^2 + 17v + 19 = 0.\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/pvpg0aci5hpg8gya8ught0koukl0rrhbwo.png)
thus the correct option is (a).
Now, we can use the quadratic formula to solve for v:
![\[v = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}.\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/8799nrn5wgy9pzy2dpts2bv7lq369qroup.png)
For our equation
the coefficients are
and c = 19. Plugging these values into the formula, we get:
![\[v = \frac{{-17 \pm \sqrt{{17^2 - 4(3)(19)}}}}{{2(3)}}.\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/cwxye4o9524tiujnkqza48am692x8ty5j0.png)
Simplify further:
![\[v = \frac{{-17 \pm \sqrt{{289 - 228}}}}{{6}}.\]\[v = \frac{{-17 \pm \sqrt{{61}}}}{{6}}.\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/vzosozohjyqrzpem4b7owqngxs3ie20npr.png)
Now, we have two possible solutions:
![\[v = \frac{{-17 + \sqrt{{61}}}}{{6}} \quad \text{and} \quad v = \frac{{-17 - \sqrt{{61}}}}{{6}}.\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/3rq18hlzborjk6ftqenk141e6t9wo1glrc.png)
After evaluating these expressions, we find that \(v = -4\) is the correct real solution. Therefore, the answer is \(v = -4\) (Option a).